If x neq 0,then frac{sin (pi+x) cos (frac{pi} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the allied angle formulas: $\sin(\pi+x) = -\sin x$ $\cos(\frac{\pi}{2}+x) = -\sin x$ $	an(\frac{3\pi}{2}-x) = \cot x$ $\cot(2\pi-x) = -\cot

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Using the allied angle formulas: $\sin(\pi+x) = -\sin x$ $\cos(\frac{\pi}{2}+x) = -\sin x$ $	an(\frac{3\pi}{2}-x) = \cot x$ $\cot(2\pi-x) = -\cot x$ $\sin(2\pi-x) = -\sin x$ $\cos(2\pi+x) = \cos x$ $\operatorname{cosec}(-x) = -\operatorname{cosec} x$ $\sin(\frac{3\pi}{2}+x) = -\cos x$ Substituting these values into the expression: $= \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}$ $= \frac{-\sin^2 x \cot^2 x}{\sin x \cos x \operatorname{cosec} x \cos x}$ $= \frac{-\sin^2 x \cot^2 x}{\sin x \cos^2 x \cdot \frac{1}{\sin x}}$ $= \frac{-\sin^2 x \cot^2 x}{\cos^2 x} = -	an^2 x \cdot \cot^2 x = -1$ Wait,let us re-evaluate the signs carefully: Numerator: $(-\sin x)(-\sin x)(\cot x)(-\cot x) = -\sin^2 x \cot^2 x$ Denominator: $(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x) = -\sin x \cos^2 x \operatorname{cosec} x = -\cos^2 x$ Result: $\frac{-\sin^2 x \cot^2 x}{-\cos^2 x} = \frac{\sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x}}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1$ Thus,the correct option is $C$.

If $x \neq 0$,then $\frac{\sin (\pi+x) \cos (\frac{\pi}{2}+x) \tan (\frac{3 \pi}{2}-x) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \operatorname{cosec}(-x) \sin (\frac{3 \pi}{2}+x)} = $

Similar Questions

Find the values of the other five trigonometric functions if $\cot x = \frac{3}{4}$ and $x$ lies in the third quadrant.

The value of $\tan 67 \frac{1}{2}^{\circ} + \cot 67 \frac{1}{2}^{\circ}$ is

If $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$,then $a^2+b^2=$

The value of $\tan \frac{\pi}{3} + 2 \tan \frac{2 \pi}{3} + 4 \tan \frac{4 \pi}{3} + 8 \tan \frac{8 \pi}{3}$ is equal to (in $\sqrt{3}$)

If $\theta$ lies in the second quadrant,then the value of $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module